Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100 + 5;
const int INF = 1e9;
vector<vector<int> > a(N, vector<int>(N)), b(N, vector<int>(N));
bool solve(vector<vector<int> >& a, vector<string>& ans1, vector<int>& ans2,
           int n, int m) {
  for (int i = 1; i <= n; ++i) {
    int curmin = INF;
    for (int j = 1; j <= m; ++j) curmin = min(curmin, a[i][j]);
    for (int j = 1; j <= m; ++j) a[i][j] -= curmin;
    while (curmin--) {
      ans1.push_back("row");
      ans2.push_back(i);
    }
  }
  for (int j = 1; j <= m; ++j) {
    int curmin = INF;
    for (int i = 1; i <= n; ++i) curmin = min(curmin, a[i][j]);
    for (int i = 1; i <= n; ++i) a[i][j] -= curmin;
    while (curmin--) {
      ans1.push_back("col");
      ans2.push_back(j);
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (a[i][j] != 0) {
        return false;
      }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) cin >> a[i][j], b[j][i] = a[i][j];
  vector<string> anss1;
  vector<int> ans1;
  vector<string> anss2;
  vector<int> ans2;
  if (solve(a, anss1, ans1, n, m) && solve(b, anss2, ans2, m, n)) {
    if (anss1.size() < anss2.size()) {
      cout << anss1.size() << '\n';
      for (int i = 0; i < anss1.size(); ++i)
        cout << anss1[i] << ' ' << ans1[i] << '\n';
    } else {
      cout << anss2.size() << '\n';
      for (int i = 0; i < anss2.size(); ++i) {
        if (anss2[i] == "row")
          cout << "col ";
        else
          cout << "row ";
        cout << ans2[i] << '\n';
      }
    }
  } else
    cout << -1;
  return 0;
}